# Chess Set Pricing and Profit Problem Given: - **Demand Function:** \( q(p) = -5.5p + 1400 \), where \( q \) is the number of sets sold per month at price \( p \). - **Domain:** \( 0 \leq p \leq 254.55 \) - **Cost per set:** $75 - **Monthly markup:** 45% Let's address each part step-by-step. --- ## (a) Model for Monthly Profit Let: - \( p \): **Wholesale price per set** - \( C \): **Cost per set** (\$75) - \( q(p) \): **Number of sets sold at price \( p \)** **Profit per set:** \( p - C \) **Total monthly profit:** \[ \text{Profit}(p) = (\text{Profit per set}) \times (\text{Number of sets sold}) \] \[ \boxed{ \Pi(p) = (p - 75)(-5.5p + 1400) } \] **Expanded form:** \[ \Pi(p) = (p - 75)(-5.5p + 1400) \] \[ = -5.5p^2 + 1400p + 412.5p - 105,000 \] \[ = -5.5p^2 + 1812.5p - 105,000 \] --- ## (b) Monthly Profit if Wholesale Price is \$100 Substitute \( p = 100 \) into the profit function: \[ \Pi(100) = (100 - 75)(-5.5 \times 100 + 1400) \] \[ = 25 \times (-550 + 1400) \] \[ = 25 \times 850 = \boxed{\$21,250} \] **Monthly profit is \$21,250 if the wholesale price is \$100.** --- ## (c) Wholesale Price for Monthly Profit of \$8,000 Set profit function equal to \$8,000 and solve for \( p \): \[ (p - 75)(-5.5p + 1400) = 8000 \] \[ -5.5p^2 + 1400p + 412.5p - 105,000 = 8000 \] \[ -5.5p^2 + 1812.5p - 105,000 = 8000 \] \[ -5.5p^2 + 1812.5p - 113,000 = 0 \] Multiply both sides by \(-1\): \[ 5.5p^2 - 1812.5p + 113,000 = 0 \] This is a quadratic equation: \( 5.5p^2 - 1812.5p + 113,000 = 0 \) ### Use quadratic formula: \[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 5.5 \), \( b = -1812.5 \), \( c = 113,000 \): \[ p = \frac{1812.5 \pm \sqrt{(-1812.5)^2 - 4 \times 5.5 \times 113,000}}{2 \times 5.5} \] Calculate under the square root: \[ (-1812.5)^2 = 3,286,156.25 \] \[ 4 \times 5.5 \times 113,000 = 2,486,000 \] \[ \sqrt{3,286,156.25 - 2,486,000} = \sqrt{800,156.25} \approx 894.5 \] So, \[ p = \frac{1812.5 \pm 894.5}{11} \] Calculate both solutions: - \( p_1 = \frac{1812.5 + 894.5}{11} = \frac{2707}{11} \approx 246.09 \) - \( p_2 = \frac{1812.5 - 894.5}{11} = \frac{918}{11} \approx 83.45 \) **Check which is in the valid domain** \( 0 \leq p \leq 254.55 \): - Both are in the domain, but for a practical wholesale price (markup is 45% over cost, so typical price is around \$109), the lower price is likely correct. **Final Answer:** \[ \boxed{ p \approx \$83.45 } \] --- ## **Summary Table** | Part | Question | Answer/Formula | |------|------------------------------------------------------|---------------------------------------------| | (a) | Profit function | \(\Pi(p) = (p - 75)(-5.5p + 1400)\) | | (b) | Profit at \(p = \$100\) | \$21,250 | | (c) | Price for \$8,000 profit | \$83.45 (rounded to two decimals) | --- ## **Explanation in Short** - **Profit is maximized by balancing price and sales volume.** - **Lower prices increase sales but decrease profit per item, and vice versa.** - **To target a specific profit, solve the quadratic equation for \( p \).**